第3.4讲复变函数解析函数

1、( )f zwD0zD0,zzD 000()( )limzf zzf zz ( )f zw0z()f z0z0()fz0d|dzzzw00000()( )d|( ) limdz zzzf zzf zf zzz wz0y zx 00()( )limlim1zxf zzf zxzx z0 x izy 00()( )ilimlim1izyf zzf zyzy ( )f zzzzfRe)( )f zw0z000()( )( )()f zzf zf zzz z w0lim () 0zz () zz z0( )f zz( )f zw0()fzz( )f zw0d( )f zzw0z0zw( )f zz内可

2、导在内解析在DzfDzf)()(内可导在内解析在ozzfzzf)()(0连续在ozzf)( 0,0yzx 00()( )( ) lim(, )( , ) i (, )( , ) lim =izxf zzf zf zzu xx yu x yxx yx yxuxx vvv0,i0 xzy 00()( )( )lim( ,)( , ) i ( ,)( , ) limi1 =iizyf zzf zf zzu x yyu x yx yyx yyuuyyyy vvvv两者应该相等，故有 iiuuxxyy vv, uuxyxy vv, xyxyuuvv( )( , ) i ( , )f zu x yx yw

3、vizxy( , )x y, uuxyxy vv处可微（在点),),(),(yxyxvyxu( )( , ) i ( , )f zu x yx ywv( )( , )i ( , )f zu x yx ywv),),yxyx（和和（),),yxyx（和和（)(zf( )f zz, u xyv1, 1, 0, 0 xyyxuu vvxyu v2Re( )izzxxyw2,uxxyv2 0uxxxyuyxyvv0 xyRe( )zzw0zizxyi(cosisin )zxyxeeeyyz0y ziyicosisinyeyy| Arg( )2 zxzeeeyk(4.3.2)(4.3.3)kze0ze

4、1212zzzzeee2ize2 iki2 i2 zkzkzeeeekxe0z zewwLnzwii , uz re wviiuerevii(2 )ukerevln , 2 Argurkz vArgzzizzarglnlnikzLnz2ln 为为整整数数kxzLnxzlnln0就就是是通通常常的的对对数数主主值值时时，特特别别地地，当当单值单值多多值值且且是是无无穷穷多多值值的的对对数数，）：负负数数在在复复数数范范围围有有说说明明（ 11n LnznzLnLnznzLnnn1122(1i)2zz3i 412zze123lnlni4zz-i21 2ln()ln( i)lni2zze 1212l

5、n()lnlnz zzz.不不成成立立的的说说明明：在在主主值值范范围围内内是是LnzzLn21LnznzLnLnznzLnnn1lnziz x y 221lnln| | iargln() iarg2zzzxyz221lnln()2zxyargzarg zyiicosisin ,cosisin .yyeyyeyyiiiisin, cos 2 i2yyyyeeeeyyiiiisin, cos2i2zzzzeeeezz”改改为为复复数数“如如果果将将实实数数“zysin , coszz2ze2i sin(2) sin ,cos(2) coszzzzsinzsin()sinzz cos zcos()

6、coszzsin , coszz(sin )coszz (cos )sinzz icosisinzezz,cosz)e(e212i1)2-e()siniz-iz-izieieieziziziz（证证明明：121212sin() sin coscos sinzzzzzz121212cos()coscossinsinzzzzzz22sincos1zz sin1costan, cot =,costanzsinzzzzzztanztgzcotzctgzzcoszzarccos)-1(-arccos2ziziLnzziziLniziziLnz-2-arctan)-1(-arcsin2)32(ln-2i2

7、k)3ln(2-)32(-2arccosikiiLn解：解：2arccos为整数）为整数）k(zLnzezz0z0z22( , )u x yxyxy( )( , ) i ( , )f zu x yx y v(0) 0f( , )u x y2, 2xyuxyuxy2, 20 xxyyxxyyuuuu( , )u x y, x yv2yxuxyv21(2)d2( )2x y yxyyxv21( )2xxc 22( , )22yxx yxycv222211( )ii 2 iii22f zuxyxyyxyxc v(0) 0f0c 222211( )(i )i(2i)(1i)22f zxyxyxyz ( , )u x y( )fz( )ii(2) i(2)2(i ) i(i ) 2i(2 i)xxxyf zuuux yy xxyxyzzzv2i( )(2 i) d(1)2f zz zzc(0)0f21( )(1i)2f zz( ,)u x y2, 2yxxyux y